python笔记——多个decorator的执行顺序

事实上，用这样一段代码就能看出执行顺序：

def decorator_a(func):
    print('decorator a running.')

    def deco_a(*args, **kwargs):
        print('pretreatment a')
        res = func(*args, **kwargs)
        print('post processing a')
        return res

    print('decorator a finished.')
    return deco_a


def decorator_b(func):
    print('decorator b running.')

    def deco_a(*args, **kwargs):
        print('pretreatment b')
        res = func(*args, **kwargs)
        print('post processing b')
        return res

    print('decorator b finished.')
    return deco_a


@decorator_a
@decorator_b
def f(x):
    print('running f.')
    return x*2


print('')


def g(x):
    print('running g.')
    return x*3


g = decorator_a(g)
g = decorator_b(g)


print('')
print('\n', f(1), '\n')
print('\n', g(1), '\n')

输出：

decorator b running.
decorator b finished.
decorator a running.
decorator a finished.

decorator a running.
decorator a finished.
decorator b running.
decorator b finished.

pretreatment a
pretreatment b
running f
post processing b
post processing a

 2 

pretreatment b
pretreatment a
get in g
post processing a
post processing b

 3

首先，嵌套函数的外层内容在函数定义并装饰的过程中就已经被执行了。其次，装饰器的装饰顺序是靠近函数的装饰器先进行装饰，远离函数定义位置的装饰器后进行装饰。装饰函数内定义的函数定义之前的代码会在装饰过程中执行，而内层函数则会被返回给原函数。换而言之，在第一次装饰（@decorator_b）后，f已经变成了deco_b。而在第二次装饰时，相当于deco_b被传入decorator_a。装饰过程等于下列装饰函数调用

def i(x):
    return x*2


i = decorator_a(decorator_b(i))
print('')
print(i(1))

上述例子中，显示的，装饰过程也可以等价于下列过程：

print('decorator b running.')
print('decorator b finished.')
print('decorator a running.')
print('decorator a finished.')


def h(x):
    print('pretreatment a')
    print('pretreatment b')
    print('running h.')
    res = x*2
    print('post processing b')
    print('post processing a')
    return res


print(h(1), '\n')

而调用函数进行装饰则正好相反，调用过程是自上往下的，调用过程必须写在函数下方。

python笔记——多个decorator的执行顺序

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